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Question 4.1: The function $f$ is periodic with period $T=4$, defined as: $f(x) = \begin{cases} x^2 - 4 & \text{if } 0 \le x < 2 \\ b - ax & \text{if } 2 \le x < 4 \end{cases}$ It is given that $f$ is continuous at $x=2$. For a periodic function to be continuous over its entire domain, it must also satisfy $f(0) = \lim_{x \to 4^-} f(x)$.

a) The value of the constant $b$.

Step 1: Use the continuity at $x=2$. For $f(x)$ to be continuous at $x=2$, the left-hand limit must equal the right-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)$ Using the first piece for $x < 2$: $\lim_{x \to 2^-} (x^2 - 4) = 2^2 - 4 = 4 - 4 = 0$ Using the second piece for $x \ge 2$: $\lim_{x \to 2^+} (b - ax) = b - a(2) = b - 2a$ Equating these, we get our first equation: $b - 2a = 0 \implies b = 2a \quad (*)$

Step 2: Use the continuity condition for the periodic function at the interval boundaries. For a periodic function defined on $[0, T)$ to be continuous, $f(0)$ must equal $\lim_{x \to T^-} f(x)$. Here $T=4$. $f(0) = 0^2 - 4 = -4$ $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} (b - ax) = b - a(4) = b - 4a$ Equating these, we get our second equation: $b - 4a = -4 \quad (**)$

Step 3: Solve the system of equations for $a$ and $b$. Substitute $b = 2a$ from $(*)$ into $(**)$: $2a - 4a = -4$ $-2a = -4$ $a = 2$ Now substitute $a=2$ back into $b=2a$: $b = 2(2) = 4$ The value of the constant $b$ is $\boxed{4}$.

b) $f(13)$ and $f(-1.1)$.

Step 1: Calculate $f(13)$ using the periodicity. Since the period is $T=4$, $f(x+4k) = f(x)$ for any integer $k$. We find the equivalent value in the interval $[0, 4)$: $13 = 3 \times 4 + 1$ So, $f(13) = f(1)$. Since $0 \le 1 < 2$, we use $f(x) = x^2 - 4$: $f(1) = 1^2 - 4 = 1 - 4 = -3$ Therefore, $f(13) = \boxed{-3}$.

Step 2: Calculate $f(-1.1)$ using the periodicity. We find the equivalent value in the interval $[0, 4)$: $-1.1 + 1 \times 4 = 2.9$ So, $f(-1.1) = f(2.9)$. Since $2 \le 2.9 < 4$, we use $f(x) = b - ax$. With $a=2$ and $b=4$, this is $f(x) = 4 - 2x$: $f(2.9) = 4 - 2(2.9) = 4 - 5.8 = -1.8$ Therefore, $f(-1.1) = \boxed{-1.8}$.

c) Investigate the parity of $f$ in the interval $0 \le x < 2$.

Parity (even or odd) is typically defined for functions whose domain is symmetric about the origin (e.g., $[-L, L]$). The interval $[0, 2)$ is not symmetric. However, if we consider the expression $f(x) = x^2 - 4$ for $x \in [0, 2)$, this expression itself represents an even function because $g(-x) = (-x)^2 - 4 = x^2 - 4 = g(x)$. But the function $f(x)$ as a whole, considering its periodic nature, is neither even nor odd. For example, $f(1) = -3$. If $f$ were even, $f(-1)$ would be $-3$. If $f$ were odd, $f(-1)$ would be $3$. Using periodicity, $f(-1) = f(-1+4) = f(3)$. From part (a), $f(x) = 4-2x$ for $2 \le x < 4$. So, $f(3) = 4 - 2(3) = 4 - 6 = -2$. Since $f(-1) = -2$, and $f(1) = -3$: $f(-1) \ne f(1)$ (so not even). $f(-1) \ne -f(1)$ (since $-(-3) = 3$, so not odd). Therefore, the function $f(x)$ is neither even nor odd.

d) Sketch the graph of $f(x)$ in the interval $-4 \le x < 8$.

The function is $f(x) = \begin{cases} x^2 - 4 & \text{if } 0 \le x < 2 \\ 4 - 2x & \text{if } 2 \le x < 4 \end{cases}$ with period 4. We need to sketch it over three periods: $[-4, 0)$, $[0, 4)$, and $[4, 8)$.

For the interval $[0, 4)$:

• For $0 \le x < 2$, $f(x) = x^2 - 4$. This is a parabola. $f(0) = -4$ $f(1) = -3$ $f(2) = 0$ (approaching from left)
• For $2 \le x < 4$, $f(x) = 4 - 2x$. This is a straight line. $f(2) = 0$ $f(3) = -2$ $f(4) = -4$ (approaching from left, $f(4)$ is technically $f(0)$ due to periodicity)

For the interval $[4, 8)$: Using $f(x) = f(x-4)$:

• For $4 \le x < 6$ (i.e., $0 \le x-4 < 2$), $f(x) = (x-4)^2 - 4$. $f(4) = -4$ $f(5) = -3$ $f(6) = 0$
• For $6 \le x < 8$ (i.e., $2 \le x-4 < 4$), $f(x) = 4 - 2(x-4)$. $f(6) = 0$ $f(7) = -2$ $f(8) = -4$ (approaching from left)

For the interval $[-4, 0)$: Using $f(x) = f(x+4)$:

• For $-4 \le x < -2$ (i.e., $0 \le x+4 < 2$), $f(x) = (x+4)^2 - 4$. $f(-4) = -4$ $f(-3) = -3$ $f(-2) = 0$
• For $-2 \le x < 0$ (i.e., $2 \le x+4 < 4$), $f(x) = 4 - 2(x+4)$. $f(-2) = 0$ $f(-1) = -2$ $f(0) = -4$ (approaching from left)

The graph will consist of three identical segments. Each segment starts at $y=-4$, curves up to $y=0$, then goes down linearly to $y=-4$.

(A detailed sketch cannot be provided in this text format, but the description above outlines the shape and key points for plotting.)

Question ii) Prove by contradiction that $\frac{1}{\sqrt{2}}$ is irrational.

Step 1: Assume the opposite (for contradiction). Assume that $\frac{1}{\sqrt{2}}$ is a rational number. If $\frac{1}{\sqrt{2}}$ is rational, then it can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers, $q \ne 0$, and $p$ and $q$ have no common factors (i.e., the fraction is in its simplest form). $\frac{1}{\sqrt{2}} = \frac{p}{q}$

Step 2: Manipulate the equation to isolate $\sqrt{2}$. Cross-multiply the equation: $q = p\sqrt{2}$ Divide both sides by $p$ (since $p$ must be non-zero, otherwise $\frac{1}{\sqrt{2}}=0$, which is false): $\sqrt{2} = \frac{q}{p}$

Step 3: Analyze the implication. Since $p$ and $q$ are integers and $p \ne 0$, the fraction $\frac{q}{p}$ represents a rational number. This implies that $\sqrt{2}$ is a rational number.

Step 4: State the contradiction. However, it is a well-established mathematical fact that $\sqrt{2}$ is an irrational number. The conclusion that $\sqrt{2}$ is rational directly contradicts this known fact.

Step 5: Conclude. Since our initial assumption (that $\frac{1}{\sqrt{2}}$ is rational) leads to a contradiction, the assumption must be false. Therefore, $\frac{1}{\sqrt{2}}$ is an irrational number.